Difference between revisions of "2017 AIME I Problems/Problem 7"
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==Solution== | ==Solution== | ||
Let <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to <math>\binom{18}{6}=18564</math>. Therefore, the answer is <math>\boxed{564}</math>. | Let <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to <math>\binom{18}{6}=18564</math>. Therefore, the answer is <math>\boxed{564}</math>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=6|num-a=8}} | {{AIME box|year=2017|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:18, 8 March 2017
Problem 7
For nonnegative integers and
with
, let
. Let
denote the sum of all
, where
and
are nonnegative integers with
. Find the remainder when
is divided by
.
Solution
Let , and note that
. The problem thus asks for the sum
over all
such that
. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to
. Therefore, the answer is
.
-rocketscience
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.